Problems

Solution: While descending, the body experiences the force \(-\frac15mg - 2mv^2\). \begin{align*} \text{N2:} && m \dot{v} &= -\frac15 mg - 2mv^2 \\ \Rightarrow && \frac{\dot{v}}{\frac15g + 2v^2} &= -1 \\ \Rightarrow && \frac{1}{2}\tan^{-1} v_1 - \frac{1}{2}\tan^{-1} {v_0} &= -T \end{align*} We care about when \(v_1 = 0\), ie \(\displaystyle T = \frac{1}{2}\tan^{-1} {v_0} < \frac12 \frac{\pi}2 = \frac{\pi}4\) seconds. If the body enters at speed \(1\ \mathrm{ms}^{-1}.\) then for the first part of it's journey it will experience forces \(-\frac15mg - 2mv^2\) and so: \begin{align*} \text{N2:} && m v \frac{\d v}{\d x} &= -\frac15 mg - 2mv^2 \\ \Rightarrow && \int \frac{v}{2(1 + v^2)} \d v &= \int -1 \d x \\ \Rightarrow && \frac14 \ln (1 + v^2) &= -x \end{align*} Therefore the depth is \(\frac14 \ln 2\) metres. When the body is rising, it experiences forces of: \(\frac15mg - 2mv^2\) and so: \begin{align*} \text{N2:} && m v \frac{\d v}{\d x} &= \frac15mg - 2mv^2 \\ \Rightarrow && \int \frac{v}{2(1 - v^2)} \d v &= \int -1 \d x \\ \Rightarrow && -\frac14 \ln (1 - v^2) &= \frac14 \ln 2 \\ \Rightarrow && 1-v^2 &= \frac12 \\ \Rightarrow && v &= \frac{1}{\sqrt{2}} \ \mathrm{ms}^{-1} \end{align*} This will take \begin{align*} \text{N2:} && m \dot{v} &= \frac15mg - 2mv^2 \\ \Rightarrow && \frac{\dot{v}}{2(1-v^2)} &= -1 \\ \Rightarrow && \dot{v} \frac{1}{4}\l \frac{1}{1 - v} + \frac{1}{1+v} \r &= -1 \\ \Rightarrow && \frac{1}{4} \l -\ln(1 - v) + \ln(1 + v)\r &= -T \end{align*} Since \(v = \frac{1}{\sqrt{2}}\) \begin{align*} T &= \frac{1}{4} \ln \l \frac{1+ \frac1{\sqrt{2}}}{1 - \frac1{\sqrt{2}}}\r \\ &= \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \end{align*} and therefore the total time will be: \begin{align*} & \frac12 \tan^{-1} 1 + \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \\ =& \frac{\pi}{8} + \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \end{align*}

Solution: Consider \(E = \frac12 m \dot{x}^2 - \frac{Gm}{x}\), notice that \begin{align*} && \dot{E} &= m \dot{x} \ddot{x} + \frac{Gm}{x^2} \dot{x} \\ &&&= \dot{x} \underbrace{\left (m\ddot{x} + \frac{Gm}{x^2} \right)}_{=0 \text{ by N2}} \end{align*} Therefore \(E\) is conserved. Therefore if \(x \to \infty\) \(\frac12 m V^2 - \frac{Gm}{h} = \frac12 m u^2 - 0 \geq 0\) so \(V^2 \geqslant 2G/h\) Since \(E \approx 0\) we want to solve \begin{align*} && \dot{x} &= -\sqrt{\frac{2G}{x}} \\ \Rightarrow && -\int_h^0 \sqrt{x} \d x &= \int_0^T \sqrt{2G} \d t \\ \Rightarrow && \frac{2h^{3/2}}{3} &= \sqrt{2G}T \\ \Rightarrow && T &= \frac{\sqrt{2}h^{3/2}}{3\sqrt{G}} = \frac13 \sqrt{\frac{2h^3}{G}} \end{align*}

Solution: \begin{align*} && y & =x\frac{\mathrm{d}y}{\mathrm{d}x}-\cosh\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{\d y}{\d x} + x\frac{\d ^2 y}{\d x^2} - \sinh \left ( \frac{\d y}{\d x} \right) \frac{\d^2 y}{\d x^2} \\ \Rightarrow && 0 &= \frac{\d^2 y}{\d x^2} \left ( x - \sinh \left ( \frac{\d y}{\d x}\right)\right) \end{align*} Therefore \(\frac{\d^2y}{\d x^2} = 0\) or \( x - \sinh \left ( \frac{\d y}{\d x}\right) = 0\) as required. \begin{align*} && \frac{\d ^2 y}{\d x^2} &= 0 \\ \Rightarrow && y &= ax + b \\ \\ && 0 &= x - \sinh \left ( \frac{\d y}{\d x}\right) \\ \Rightarrow && \frac{\d y}{\d x} &= \sinh^{-1} (x) \\ \Rightarrow && y &= x \sinh^{-1} x - \sqrt{x^2+1} + C \end{align*} Since it is necessary the solution satisfies one of those equations, we just need to check if either of these types of solutions work for our differential equation, ie \begin{align*} && ax + b &\stackrel{?}{=} ax - \cosh(a) \\ \Rightarrow && b &= -\cosh(a) \\ \Rightarrow && y &= ax -\cosh(a) \\ \\ && x \sinh^{-1} x - \sqrt{x^2+1} + C &\stackrel{?}{=} x\sinh^{-1} x - \cosh ( \sinh^{-1} x) \\ &&&= \sinh^{-1} x -\sqrt{x^2+1} \\ \Rightarrow && C &= 0 \end{align*} Therefore the general solutions are, \(y = ax - \cosh(a)\) and \(y = x \sinh^{-1} x - \sqrt{x^2+1}\)

Solution: Note that \(z = xy/(x+y+1) \Rightarrow y(x-z) = z(x+1)\) \begin{align*} && g(x) + g(y) &= g(z) \\ \Rightarrow && g'(x) &= g'(z) \cdot \frac{y(x+y+1) - xy \cdot 1} {(x+y+1)^2} \\ &&&= g'(z) \frac{y^2+y}{(x+y+1)^2} \\ &&&= g'(z) \frac{z^2(y^2+y)}{x^2y^2} \\ &&&= g'(z) \frac{z^2(y+1)}{x^2y} \\ &&&= g'(z) \frac{z^2}{x^2} \left (1 + \frac{x-z}{z(x+1)} \right) \\ &&&= g'(z) \frac{z}{x^2} \frac{zx+x}{x+1} \\ &&&= g'(z) \frac{z(z+1)}{x(x+1)} \end{align*} If \(x = 1\) then as \(y\) ranges from \(0\) to \(\infty\), \(z\) ranges from \(0\) to \(1\), so \(g'(1) = \frac{z(z+1)}{1(1+1)}g'(z)\), ie \(2g'(1) = (u^2+u)g'(u)\). \begin{align*} && g'(u) &= \frac{A}{u(u+1)} \\ \Rightarrow && g(u) &= A\int \left ( \frac{1}{u} - \frac{1}{u+1} \right) \d u \\ &&&= A \left ( \ln u - \ln(u+1) \right) + B \\ &&&= A \ln \left ( \frac{u}{u+1} \right) + B \\ \\ && A \ln \left ( \frac{x}{x+1} \right) + B+A \ln \left ( \frac{y}{y+1} \right) + B &=A \ln \left ( \frac{z}{z+1} \right) + B \\ \Rightarrow && B &= A \ln \left ( \frac{z}{z+1} \frac{y+1}{y} \frac{x+1}{x} \right) \\ &&&= A \ln \left ( \frac{1}{1+\frac{x+y+1}{xy}} \frac{(y+1)(x+1)}{xy} \right) \\ &&&= A \ln 1 \\ &&& = 0 \end{align*} Therefore \(B = 0\). \(A\) cannot be determined from \((*)\). Suppose \(f(x) + f(y) = f(z)\), then \(f'(x) = yf'(z)\). Letting \(x = 1\) we find \(f'(1) = uf'(u) \Rightarrow f(u) = C \ln u + D\), but \(D = 0\) so \(f(x) = C \ln x\)

Solution: \begin{align*} && 0 &= u^2 + 2u \sinh x -1 \\ &&&= u^2 + u(e^x-e^{-x})-e^{x}e^{-x} \\ &&&= (u-e^{-x})(u+e^x) \\ \Rightarrow && u &= e^{-x}, -e^x \end{align*} \begin{align*} && 0 &= \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}\sinh x-1 \\ \Rightarrow && \frac{\d y}{\d x} &= e^{-x}, -e^x \\ \Rightarrow && y &= -e^{-x}+C, -e^x+C \\ y(0) = 0: && C &= 1\text{ both cases } \\ y'(0) > 0: && y &= 1-e^{-x} \end{align*} \begin{align*} && 0 &= \sinh x u^2 + 2u -\sinh x \\ \Rightarrow && u &= \frac{-2 \pm \sqrt{4+4\sinh^2 x}}{2\sinh x} \\ &&&= \frac{-1 \pm \cosh x}{\sinh x} = - \textrm{cosech }x \pm \textrm{coth}x \\ \\ && 0 &= \sinh x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}-\sinh x \\ \Rightarrow && \frac{\d y}{\d x} &= - \textrm{cosech }x \pm \textrm{coth}x \\ \Rightarrow && y &= -\ln \left ( \tanh \frac{x}{2} \right) \pm \ln \sinh x+C \end{align*} For \(x \to 0\) to be defined, we need \(+\), so \begin{align*} && y &= \ln \left (\frac{\sinh x}{\tanh \frac{x}{2}} \right) + C \\ && y &= \ln \left (\frac{2\sinh \frac{x}{2} \cosh \frac{x}{2}}{\tanh \frac{x}{2}} \right)+C \\ &&&= \ln \left (2 \cosh^2 x \right) + C \\ y(0) = 0: && 0 &= \ln 2+C \\ \Rightarrow && y &= \ln(2 \cosh^2 x) -\ln 2 \\ && y &= 2 \ln (\cosh x) \end{align*}

Solution: First we seek the complementary function. \begin{align*} && \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r &= 0 \\ \Rightarrow && r &= A \sin 2\theta + B \cos 2 \theta \end{align*} Next we seek a particular integral, of the form \(r = C \sin 3 \theta\). \begin{align*} && \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r &= 5 \sin 3 \theta \\ \Rightarrow && -9C \sin 3 \theta + 4C \sin 3 \theta &= 5 \sin 3 \theta \\ \Rightarrow && C &= -1 \\ \end{align*} So our general solution is \(A \sin 2\theta + B \cos 2 \theta -\sin 3 \theta\). Plugging in boundary conditions we obtain: \begin{align*} \theta = \frac{\pi}{2}, r = 1: &&1 &= -B +1 \\ \Rightarrow && B &= 0 \\ \theta = \frac{\pi}{2}, \frac{\d r}{\d \theta} = -2: && -2 &= -2A \\ \Rightarrow && A &= 1 \end{align*} So the general solution is \(r = \sin 2 \theta - \sin 3 \theta = 2 \sin \left ( \frac{-\theta}{2} \right) \cos \left (\frac{5 \theta}{2} \right)\) First notice that for \(\theta \in \left [\frac{\pi}{5}, \frac{3 \pi}{5} \right]\) this is positive, and it is zero on the end points, therefore we are tracing out a a loop. The area of the loop will be: \begin{align*} A &= \int_{\pi/5}^{3\pi/5} \frac12 \left ( \sin 2 \theta - \sin 3 \theta \right)^2 \d \theta \\ &= \frac12\int_{\pi/5}^{3\pi/5} \sin^2 2\theta + \sin^2 3 \theta - 2 \sin 2 \theta \cos 3 \theta \d \theta \\ &= \frac12\int_{\pi/5}^{3\pi/5} \frac{1-2 \cos 4 \theta}{2} + \frac{1-2 \cos6 \theta}{2} - \sin5 \theta-\cos\theta \d \theta \\ &= \frac12 \left [\theta - \frac14 \sin 4 \theta-\frac16 \sin 6 \theta + \frac15 \cos 5 \theta - \sin \theta \right]_{\pi/5}^{3\pi/5} \\ &= \frac{\pi}{5} +\frac{25}{48}\left [ \sin\left(\frac{\pi}{5}\right)-\sin\left(\frac{2\pi}{5}\right) \right] \end{align*}

Solution: Suppose \(y = uv\) then and suppose \(\frac{\d u}{\d x} + P u = 0\) then \begin{align*} && \frac{\d y}{\d x} + Py &= Q \\ && uv' + u'v + Puv &= Q \\ && uv' &= Q \\ && \frac{\d v}{\d x} &= \frac{Q}{u} \end{align*} Consider \begin{align*} && 0 &= \frac{\d u}{\d x} - \frac{2u}{x+1} \\ \Rightarrow && \ln u &= 2\ln (1 + x) + C \\ \Rightarrow && u &= A(1+x)^2 \end{align*} and \begin{align*} && \frac{\d v}{\d x} &= \frac1{A}(x+1)^{\frac12} \\ \Rightarrow && v &= \frac2{3A}(x+1)^{\frac32} + k \\ \Rightarrow && y &= \frac23(x+1)^\frac72 + k(x+1)^2 \\ && 0 &= y(1) \\ &&&= \frac23 2^{7/2}+k2^2 \\ \Rightarrow && k &= -\frac{2^{5/2}}{3} \\ \Rightarrow && y &= \frac23 (x+1)^{7/2} - \frac{2^{5/2}}{3}(x+1)^2 \end{align*}

Solution: Suppose \(f(t) = Ae^{pt}\) is a solution, then \begin{align*} && 0 &= Ape^{pt} + Ae^{pt} - Ake^{p(t-1)} \\ \Leftrightarrow && 0 &= p +1 - ke^{-p} \\ \Leftrightarrow && p+1 &= ke^{-p} \end{align*}

+ - ↺

Solution: Consider \(\frac12 x(t)^2\) then differentiating we obtain \(x(t)x'(t) = x(t)y(t)\). Also note that \(x(0) = \int_0^0 y(u) \d u = 0\) Therefore, \begin{align*} \int_0^1 x(t)y(t) \d t &= \left [ \frac12 x(t)^2 \right]_0^1 \\ &= \frac12[x(1)]^2 \end{align*} \begin{align*} && 0 &= y'' + (y^2-1)y' + y \\ \Rightarrow && 0 &= \int_0^1 \l y'' + (y^2-1)y' + y \r \d t \\ &&&= \left [y'(t) + \frac13y^3-y+x \right]_0^1 \\ &&&= x(1) \end{align*} Therefore \(x(1) = 0\). \begin{align*} && 0 &= xy'' + (y^2-1)y' x+ yx \\ \Rightarrow && 0 &= \int_0^1 \l xy'' + (y^2-1)y'x + xy \r \d t \\ &&&= \left [ x y' +(\frac13 y^3-y)x \right]_0^1 - \int_0^1 yy'+\frac13y^4-y^2 \d t \\ &&&= 0 - \frac13 \int_0^1 [y(t)]^4 \d t - \int_0^1 [y(t)]^2 \d t \\ \Rightarrow && \int_0^1 [y(t)]^2 \d t &= \frac13 \int_0^1 [y(t)]^4 \d t \end{align*} \begin{align*} && 0 &= yy'' + (y^2-1)y' y+ y^2 \\ \Rightarrow && 0 &= \int_0^1 \l yy'' + (y^2-1)y'y + y^2 \r \d t \\ &&&= \left [ y y' +(\frac14 y^4-\frac12y^2) \right]_0^1 - \int_0^1 [y'(t)]^2 \d t + \int_0^1 y^2 \d t \\ &&&= 0 - \int_0^1 [y'(t)]^2 \d t + \int_0^1 y^2 \d t \\ \Rightarrow && \int_0^1 [y'(t)]^2 \d t &= \int_0^1 [y(t)]^2 \d t \end{align*}